Problem: Simplify the following expression: $\dfrac{40y^4}{45y^3}$ You can assume $y \neq 0$.
Explanation: $ \dfrac{40y^4}{45y^3} = \dfrac{40}{45} \cdot \dfrac{y^4}{y^3} $ To simplify $\frac{40}{45}$ , find the greatest common factor (GCD) of $40$ and $45$ $40 = 2 \cdot 2 \cdot 2 \cdot 5$ $45 = 3 \cdot 3 \cdot 5$ $ \mbox{GCD}(40, 45) = 5 $ $ \dfrac{40}{45} \cdot \dfrac{y^4}{y^3} = \dfrac{5 \cdot 8}{5 \cdot 9} \cdot \dfrac{y^4}{y^3} $ $\phantom{ \dfrac{40}{45} \cdot \dfrac{4}{3}} = \dfrac{8}{9} \cdot \dfrac{y^4}{y^3} $ $ \dfrac{y^4}{y^3} = \dfrac{y \cdot y \cdot y \cdot y}{y \cdot y \cdot y} = y $ $ \dfrac{8}{9} \cdot y = \dfrac{8y}{9} $